Por que estou recebendo tanto erro no meu solucionador de Runge Kutta Fehlberg?

Meu projeto atual é uma reprogramação de um modelo de dobramento de proteínas envolvendo a solução de milhares de ODEs em C ++. Tenho parado e começado a progredir enquanto escrevia o solucionador para rodar totalmente na GPU. Finalmente consegui integrar, mas quando tento resolver dC / dt = -C usando a solução de quinta ordem do algoritmo RKF45 com um tamanho de passo fixo de h = 0,001, obtenho uma decisão a partir do valor da calculadora de e ^ {- t } na ordem de 10 ^ {- 4} quando é esperado um erro global na ordem de 10 ^ {- 12}.

É porque não estou usando tamanhos de etapas adaptáveis ​​e controle de erros usando as soluções de quarta e quinta ordem que isso está ocorrendo? Meu pensamento era que, uma vez que um algoritmo adaptativo analisa a diferença entre as suposições de quarta e quinta ordem e trata basicamente a quinta ordem como a "resposta correta", eu usaria essa resposta para experimentar o integrador.

EDIT:

Eu pensei que desde

Erro global = (número de pontos) * (erro local)

e

Erro local = h ^ O

onde h é o tamanho da etapa e O é a ordem, eu calculo

O = ln (erro global / número de pontos) / ln h

Fazendo uma análise de pedidos da maneira que achei correta e sugerida na resposta, recebo o seguinte:

Number of Points    h       Global Error    My Way      The Answer's Way
10                  0.1     2.89E-06        6.539E+00   
20                  0.05    7.09E-08        6.495E+00   5.350E+00
40                  0.025   1.76E-07        5.216E+00   -1.315E+00
50                  0.02    1.128E-07       5.089E+00   2.003E+00
80                  0.0125  4.401E-08       4.866E+00   2.002E+00
100                 0.01    2.816E-08       4.775E+00   2.001E+00
500                 0.002   1.126E-09       4.316E+00   2.000E+00
1000                0.001   2.810E-10       4.184E+00   2.003E+00
10000               0.0001  3.000E-12       3.881E+00   1.972E+00

Portanto, apesar da minha quase inovação na noite passada, onde eu realmente tive um erro de programação, a ordem do meu solucionador ainda está desligada. O erro acima mencionado associado ao tamanho da escala 0,001 diminuiu drasticamente, no entanto, de O (10 ^ -4) para O (10 ^ -10).

2ª EDIÇÃO

Aqui está um gráfico de LN (E) -LN (a) vs. LN (h). Joguei um ajuste linear, então esperava que isso revelasse a ordem para mim. O que é realmente confuso agora é que tenho três cálculos para a ordem do meu método. O gráfico está me dizendo ~ 3, o método de Godric me diz quase exatamente 2 e meu método mostra a alteração de 6-> 4 para aumentar o tamanho da etapa. Estou mais inclinado a acreditar nos meus cálculos porque o Erro Global é muito baixo (~ 10 ^ -10) e mais próximo do que eu esperaria (10 ^ -15 * 10 ^ 3 = 10 ^ -12) para esse tamanho da etapa (10 ^ -3).

EDIT 3:

Meu código. Perdoe o noobishness.

#include <cuda.h>
#include <cuda_runtime.h>
#include <device_launch_parameters.h>
#include <stdio.h>
#include <iostream>
#include <iomanip>                      //display 2 decimal places
#include <math.h>
using namespace std;

/*
Things to do log:
1. Decide how adaptive step sizes is handled by the integrator loop.  We can no longer say how many points to integrate if that is to be determined in the kernel
itself, only what the initial step size is.
2. Figure out how I'm going to graph all this stuff.
3. On paper solve for the step size.
4. Adaptive Step Size
    a. arrMin kernel to get min of array
    b. Decide how to implement tolerances
5. Use multiples of 256 for maxlength 

Important things accomplished:
1. Can create 2D array for storing concentrations over time.
2. Can store a 1D array in 2D array.
3. Can calculate all the k-arrays given static step size.
4. Can take a step given static step size.
5. Can step through an integration from t0 to tf
6. Can accomodate fixed time-step
7. Calculated the current order of integrator using logs.
8. Learned how to profile the program.
9. Fixed the GPU timeout problem with WDDM.

Ideas:
1. Just use large fixed size array
*/
__global__ void rkf5(double*, double*, double*, double*, double*, double*, double*, double*, double*, double*, double*, int*, int*, size_t, double*, double*, double*);
__global__ void calcK(double*, double*, double*);
__global__ void k1(double*, double*, double*);
__global__ void k2(double*, double*, double*);
__global__ void k3(double*, double*, double*);
__global__ void k4(double*, double*, double*);
__global__ void k5(double*, double*, double*);
__global__ void k6(double*, double*, double*);
__global__ void arrAdd(double*, double*, double*);
__global__ void arrSub(double*, double*, double*);
__global__ void arrMult(double*, double*, double*);
__global__ void arrInit(double*, double);
__global__ void arrCopy(double*, double*);
__device__ void setup(double , double*, double*, double*, double*, int*);
__device__ double flux(int, double*) ;
__device__ double knowles_flux(int, double*);
__device__ void calcStepSize(double*, double*, double*, double*, double*, double*, double*, double*, double*, double*, double*, int*);
__global__ void storeConcs(double*, size_t, double*, int);
__global__ void takeFourthOrderStep(double*, double*, double*, double*, double*, double*, double*);
__global__ void takeFifthOrderStep(double*, double*, double*, double*, double*, double*, double*, double*);

//Error checking that I don't understand yet.
#define gpuErrchk(ans) { gpuAssert((ans), __FILE__, __LINE__); }
inline void gpuAssert(cudaError_t code, char *file, int line, bool abort=true)
{
   if (code != cudaSuccess) 
   {
      fprintf(stderr,"GPUassert: %s %s %d\n", cudaGetErrorString(code), file, line);
      if (abort) exit(code);
   }
}

//Main program.
int main(int argc, char** argv)
{
    //std::cout << std::fixed;          //display 2 decimal places
    //std::cout << std::setprecision(12);   //display 2 decimal places
    const int maxlength = 1;            //Number of discrete concentrations we are tracking.
    double concs[maxlength];            //Meant to store the current concentrations 
    double temp1[maxlength];                //Used as a bin to store products of Butcher's tableau and k values.
    double temp2[maxlength];                //Used as a bin to store products of Butcher's tableau and k values.
    double tempsum[maxlength];          //Used as a bin to store cumulative sum of tableau and k values
    double k1s[maxlength];
    double k2s[maxlength];
    double k3s[maxlength];
    double k4s[maxlength];
    double k5s[maxlength];
    double k6s[maxlength];
    const int numpoints = 40;       
    double to = 0;
    double tf = 1;
    //double dt = static_cast<double>(.5)/static_cast<double>(64);
    double dt = (tf-to)/static_cast<double>(numpoints);
    double mo = 1;
    double concStorage[maxlength][numpoints];   //Stores concs vs. time                     

    //Initialize all the arrays on the host to ensure arrays of 0's are sent to the device.
    //Also, here is where we can seed the system.
    std::cout<<dt;
    std::cout<<"\n";
    concs[0]=mo;
    std::cout<<concs[0];
    std::cout<<" ";
    for (int i=0; i<maxlength; i++)
    {
        for (int j=0; j<numpoints; j++)
            concStorage[i][j]=0;
        concs[i]=0;
        temp1[i]=0;
        temp2[i]=0;
        tempsum[i]=0;
        k1s[i]=0;
        k2s[i]=0;
        k3s[i]=0;
        k4s[i]=0;
        k5s[i]=0;
        k6s[i]=0;
        std::cout<<concs[i];
        std::cout<<" ";
    }
    concs[0]=mo;
    std::cout<<"\n";

    //Define all the pointers to device array memory addresses. These contain the on-GPU
    //addresses of all the data we're generating/using.
    double *d_concs;
    double *d_temp1;
    double *d_temp2;
    double *d_tempsum;
    double *d_k1s;
    double *d_k2s;
    double *d_k3s;
    double *d_k4s;
    double *d_k5s;
    double *d_k6s;
    double *d_dt;
    int *d_maxlength;
    int *d_numpoints;
    double *d_to;
    double *d_tf;
    double *d_concStorage;

    //Calculate all the sizes of the arrays in order to allocate the proper amount of memory on the GPU.
    size_t size_concs = sizeof(concs);
    size_t size_temp1 = sizeof(temp1);
    size_t size_temp2 = sizeof(temp2);
    size_t size_tempsum = sizeof(tempsum);
    size_t size_ks = sizeof(k1s);
    size_t size_maxlength = sizeof(maxlength);
    size_t size_numpoints = sizeof(numpoints);
    size_t size_dt = sizeof(dt);
    size_t size_to = sizeof(to);
    size_t size_tf = sizeof(tf);
    size_t h_pitch = numpoints*sizeof(double);
    size_t d_pitch;

    //Calculate the "pitch" of the 2D array.  The pitch is basically the length of a 2D array's row.  IT's larger 
    //than the actual row full of data due to hadware issues.  We thusly will use the pitch instead of the data 
    //size to traverse the array.
    gpuErrchk(cudaMallocPitch( (void**)&d_concStorage, &d_pitch, numpoints * sizeof(double), maxlength)); 

    //Allocate memory on the GPU for all the arrrays we're going to use in the integrator.
    gpuErrchk(cudaMalloc((void**)&d_concs, size_concs));
    gpuErrchk(cudaMalloc((void**)&d_temp1, size_temp1));
    gpuErrchk(cudaMalloc((void**)&d_temp2, size_temp1));
    gpuErrchk(cudaMalloc((void**)&d_tempsum, size_tempsum));
    gpuErrchk(cudaMalloc((void**)&d_k1s, size_ks));
    gpuErrchk(cudaMalloc((void**)&d_k2s, size_ks));
    gpuErrchk(cudaMalloc((void**)&d_k3s, size_ks));
    gpuErrchk(cudaMalloc((void**)&d_k4s, size_ks));
    gpuErrchk(cudaMalloc((void**)&d_k5s, size_ks));
    gpuErrchk(cudaMalloc((void**)&d_k6s, size_ks));
    gpuErrchk(cudaMalloc((void**)&d_maxlength, size_maxlength));
    gpuErrchk(cudaMalloc((void**)&d_numpoints, size_numpoints));
    gpuErrchk(cudaMalloc((void**)&d_dt, size_dt));
    gpuErrchk(cudaMalloc((void**)&d_to, size_to));
    gpuErrchk(cudaMalloc((void**)&d_tf, size_tf));

    //Copy all initial values of arrays to GPU.
    gpuErrchk(cudaMemcpy2D(d_concStorage, d_pitch, concStorage, h_pitch, numpoints*sizeof(double), maxlength, cudaMemcpyHostToDevice));
    gpuErrchk(cudaMemcpy(d_concs, &concs, size_concs, cudaMemcpyHostToDevice));
    gpuErrchk(cudaMemcpy(d_temp1, &temp1, size_temp1, cudaMemcpyHostToDevice));
    gpuErrchk(cudaMemcpy(d_temp2, &temp2, size_temp2, cudaMemcpyHostToDevice));
    gpuErrchk(cudaMemcpy(d_tempsum, &tempsum, size_tempsum, cudaMemcpyHostToDevice));
    gpuErrchk(cudaMemcpy(d_k1s, &k1s, size_ks, cudaMemcpyHostToDevice));
    gpuErrchk(cudaMemcpy(d_k2s, &k2s, size_ks, cudaMemcpyHostToDevice));
    gpuErrchk(cudaMemcpy(d_k3s, &k3s, size_ks, cudaMemcpyHostToDevice));
    gpuErrchk(cudaMemcpy(d_k4s, &k4s, size_ks, cudaMemcpyHostToDevice));
    gpuErrchk(cudaMemcpy(d_k5s, &k5s, size_ks, cudaMemcpyHostToDevice));
    gpuErrchk(cudaMemcpy(d_k6s, &k6s, size_ks, cudaMemcpyHostToDevice));
    gpuErrchk(cudaMemcpy(d_maxlength, &maxlength, size_maxlength, cudaMemcpyHostToDevice));
    gpuErrchk(cudaMemcpy(d_numpoints, &numpoints, size_numpoints, cudaMemcpyHostToDevice));
    gpuErrchk(cudaMemcpy(d_dt, &dt, size_dt, cudaMemcpyHostToDevice));
    gpuErrchk(cudaMemcpy(d_to, &to, size_to, cudaMemcpyHostToDevice));
    gpuErrchk(cudaMemcpy(d_tf, &tf, size_tf, cudaMemcpyHostToDevice));

    //Run the integrator.
    rkf5<<<1,1>>>(d_concs, d_concStorage, d_temp1, d_temp2, d_tempsum, d_k1s, d_k2s, d_k3s, d_k4s, d_k5s, d_k6s, d_maxlength, d_numpoints, d_pitch, d_dt, d_to, d_tf);
    gpuErrchk( cudaPeekAtLastError() );
    gpuErrchk( cudaDeviceSynchronize() );
    cudaDeviceSynchronize();

    //Copy concentrations from GPU to Host.  Almost defunct now that transferring the 2D array works.
    cudaMemcpy(concs, d_concs, size_concs, cudaMemcpyDeviceToHost);
    //Copy 2D array of concentrations vs. time from GPU to Host.
    gpuErrchk( cudaMemcpy2D(concStorage, h_pitch, d_concStorage, d_pitch, numpoints*sizeof(double), maxlength, cudaMemcpyDeviceToHost) );   

    //Print concentrations after the integrator kernel runs.  Used to test that data was transferring to and from GPU correctly.
    std::cout << "\n";
    for (int i=0; i<maxlength; i++)
    {
        std::cout<<concs[i];
        std::cout<<" ";
    }

    double a[10];
    double b[10];
    double c[10];
    for(int i = 0; i< 10; i++)
    {
        a[i]=0;
        b[i]=0;
        c[i]=0;
    }



    //Print out the concStorage array after the kernel runs.  Used to test that the 2D array transferred correctly from host to GPU and back.
    std::cout << "\n\n";
    std::cout << "Calculated Array";
    std::cout << "\n\n";
    for (int i=0; i<maxlength; i++)
    {
        for(int j=0; j<numpoints; j++)
        {
            if (j%(numpoints/10)==0)
            {
                a[j/(numpoints/10)]=concStorage[i][j];
                std::cout<<concStorage[i][j];
                std::cout<<"   ";
            }
        }
        std::cout << "\n";
    }
    std::cout << "\n";
    std::cout << "Exponential";
    std::cout << "\n\n";
    for (int i=0; i<10; i++)
    {
        b[i]=exp(-i*(tf-to)/10);
        std::cout<<exp(-i*(tf-to)/10);
        std::cout<<"   ";
    }
    std::cout << "\n\n";
    std::cout << "Error Array";
    std::cout << "\n\n";
    for (int i=0; i<10; i++)
    {
        c[i]=a[i]-b[i];
        std::cout<<c[i];
        std::cout<<"   ";
    }
    std::cout << "\n\n";
    cudaDeviceReset();  //Clean up all memory.

    return 0;
}
//Main kernel.  This is mean to be run as a master thread that calls all the other functions and thusly "runs" the integrator.
__global__ void rkf5(double* concs, double* concStorage, double* temp1, double* temp2, double* tempsum, double* k1s, double* k2s, double* k3s, double* k4s, double* k5s, double* k6s, int* maxlength, int* numpoints, size_t pitch, double* dt, double* to, double* tf)
{
    /*
    axy variables represent the coefficients in the Butcher's tableau where x represents the order of the step and the y value corresponds to the ky value 
    the coefficient gets multiplied by.  Have to cast them all as doubles, or the ratios evaluate as integers.
    e.g. a21 -> a21 * k1
    e.g. a31 -> a31 * k1 + a32 * k2
    */
    double a21 = static_cast<double>(.25);

    double a31 = static_cast<double>(3)/static_cast<double>(32);
    double a32 = static_cast<double>(9)/static_cast<double>(32);

    double a41 = static_cast<double>(1932)/static_cast<double>(2197);
    double a42 = static_cast<double>(-7200)/static_cast<double>(2197);
    double a43 = static_cast<double>(7296)/static_cast<double>(2197);

    double a51 = static_cast<double>(439)/static_cast<double>(216);
    double a52 = static_cast<double>(-8);
    double a53 = static_cast<double>(3680)/static_cast<double>(513);
    double a54 = static_cast<double>(-845)/static_cast<double>(4104);

    double a61 = static_cast<double>(-8)/static_cast<double>(27);
    double a62 = static_cast<double>(2);
    double a63 = static_cast<double>(-3544)/static_cast<double>(2565);
    double a64 = static_cast<double>(1859)/static_cast<double>(4104);
    double a65 = static_cast<double>(-11)/static_cast<double>(40);

    //for loop that integrates over the specified number of points. Actually, might have to make it a do-while loop for adaptive step sizes 
    for(int k = 0; k < *numpoints; k++)
    {
        if (k!=0)
        {
            arrCopy<<< 1, *maxlength >>>(concs, tempsum);
            cudaDeviceSynchronize();
        }
        arrInit<<< 1, *maxlength >>>(tempsum, 0);
        cudaDeviceSynchronize();
        arrInit<<< 1, *maxlength >>>(temp1, 0);
        cudaDeviceSynchronize();
        arrInit<<< 1, *maxlength >>>(temp2, 0);
        cudaDeviceSynchronize();

        calcK<<< 1, *maxlength >>>(concs, k1s, dt);         //k1 = dt * flux (concs)
        cudaDeviceSynchronize(); //Sync here because kernel continues onto next line before k1 finished

        setup(a21, temp1, tempsum, k1s, concs, maxlength);  //tempsum = a21*k1
        arrAdd<<< 1, *maxlength >>>(concs, tempsum, tempsum);   //tempsum = concs + a21*k1    
        cudaDeviceSynchronize();

        calcK<<< 1, *maxlength >>>(tempsum, k2s, dt);       //k2 = dt * flux (concs + a21*k1)
        cudaDeviceSynchronize();

        arrInit<<< 1, *maxlength >>>(tempsum, 0);
        cudaDeviceSynchronize();
        setup(a31, temp1, tempsum, k1s, concs, maxlength);  //temp1sum = a31*k1
        setup(a32, temp1, tempsum, k2s, concs, maxlength);  //tempsum = a31*k1 + a32*k2
        arrAdd<<< 1, *maxlength >>>(concs, tempsum, tempsum);   //tempsum = concs + a31*k1 + a32*k2
        cudaDeviceSynchronize();

        calcK<<< 1, *maxlength >>>(tempsum, k3s, dt);           //k3 = dt * flux (concs + a31*k1 + a32*k2)
        cudaDeviceSynchronize();

        arrInit<<< 1, *maxlength >>>(tempsum, 0);
        cudaDeviceSynchronize();
        setup(a41, temp1, tempsum, k1s, concs, maxlength);  //tempsum = a41*k1
        setup(a42, temp1, tempsum, k2s, concs, maxlength);  //tempsum = a41*k1 + a42*k2
        setup(a43, temp1, tempsum, k3s, concs, maxlength);  //tempsum = a41*k1 + a42*k2 + a43*k3
        arrAdd<<< 1, *maxlength >>>(concs, tempsum, tempsum);   //tempsum = concs + a41*k1 + a42*k2 + a43*k3
        cudaDeviceSynchronize();

        calcK<<< 1, *maxlength >>>(tempsum, k4s, dt);           //k4 = dt * flux (concs + a41*k1 + a42*k2 + a43*k3)
        cudaDeviceSynchronize();

        arrInit<<< 1, *maxlength >>>(tempsum, 0);
        cudaDeviceSynchronize();
        setup(a51, temp1, tempsum, k1s, concs, maxlength);  //tempsum = a51*k1
        setup(a52, temp1, tempsum, k2s, concs, maxlength);  //tempsum = a51*k1 + a52*k2
        setup(a53, temp1, tempsum, k3s, concs, maxlength);  //tempsum = a51*k1 + a52*k2 + a53*k3
        setup(a54, temp1, tempsum, k4s, concs, maxlength);  //tempsum = a51*k1 + a52*k2 + a53*k3 + a54*k4
        arrAdd<<< 1, *maxlength >>>(concs, tempsum, tempsum);   //tempsum = concs + a51*k1 + a52*k2 + a53*k3 + a54*k4
        cudaDeviceSynchronize();

        calcK<<< 1, *maxlength >>>(tempsum, k5s, dt);           //k5 = dt * flux (concs + a51*k1 + a52*k2 + a53*k3 + a54*k4)
        cudaDeviceSynchronize();

        arrInit<<< 1, *maxlength >>>(tempsum, 0);
        cudaDeviceSynchronize();
        setup(a61, temp1, tempsum, k1s, concs, maxlength);  //tempsum = a61*k1
        setup(a62, temp1, tempsum, k2s, concs, maxlength);  //tempsum = a61*k1 + a62*k2
        setup(a63, temp1, tempsum, k3s, concs, maxlength);  //tempsum = a61*k1 + a62*k2 + a63*k3
        setup(a64, temp1, tempsum, k4s, concs, maxlength);  //tempsum = a61*k1 + a62*k2 + a63*k3 + a64*k4
        setup(a65, temp1, tempsum, **k4s**, concs, maxlength);  //tempsum = a61*k1 + a62*k2 + a63*k3 + a64*k4 + a65*k5
        arrAdd<<< 1, *maxlength >>>(concs, tempsum, tempsum);   //tempsum = concs + a61*k1 + a62*k2 + a63*k3 + a64*k4 + a65*k5
        cudaDeviceSynchronize();

        calcK<<< 1, *maxlength >>>(tempsum, k6s, dt);           //k6 = dt * flux (concs + a61*k1 + a62*k2 + a63*k3 + a64*k4 + a65*k5)
        cudaDeviceSynchronize();

        //At this point, temp1 and tempsum are maxlength dimension arrays that are able to be used for other things.

        //Calculate acceptable step size before storing the concentrations.
        calcStepSize(temp1, temp2, tempsum, concs, k1s, k2s, k3s, k4s, k5s, k6s, dt, maxlength);    //temp1 = 4th Order guess, tempsum = 5th Order guess
        cudaDeviceSynchronize();

        //Store the initial conditions in the first column of the storage array.
        if (k==0)
        {
            storeConcs<<< 1, *maxlength >>>(concStorage, pitch, concs, k);  //Store this step's concentrations in 2D array
            cudaDeviceSynchronize();
        }
        //Store future concentration in next column of storage array.
        storeConcs<<< 1, *maxlength >>>(concStorage, pitch, tempsum, k+1);  //Store this step's concentrations in 2D array
        cudaDeviceSynchronize();
    }
}
//calcStepSize will take in an error tolerance, the current concentrations and the k values and calculate the resulting step size according to the following equation
//e[n+1]=y4[n+1] - y5[n+1]
__device__ void calcStepSize(double* temp1, double*temp2, double* tempsum, double* concs, double* k1s, double* k2s, double* k3s, double* k4s, double* k5s, double* k6s, double* dt, int* maxlength)
{
    //do
    //{
        takeFourthOrderStep<<< 1, *maxlength >>>(temp1, concs, k1s, k2s, k3s, k4s, k5s);            //Store 4th order guess in temp1
        takeFifthOrderStep<<< 1, *maxlength >>>(tempsum, concs, k1s, k2s, k3s, k4s, k5s, k6s);  //Store 5th order guess in tempsum
        cudaDeviceSynchronize();
        arrSub<<< 1, *maxlength >>>(temp1, tempsum, temp2)
        arrMin<<< 1, *maxlength >>>
    //arrMult
    //}
    //while
}
//takeFourthOrderStep is going to overwrite the old temp1 array with the new array of concentrations that result from a 4th order step.  This kernel is meant to be launched 
// with as many threads as there are discrete concentrations to be tracked.
__global__ void takeFourthOrderStep(double* y4, double* concs, double* k1s, double* k2s, double* k3s, double* k4s, double* k5s)
{
    double b41 = static_cast<double>(25)/static_cast<double>(216);
    double b42 = static_cast<double>(0);
    double b43 = static_cast<double>(1408)/static_cast<double>(2565);
    double b44 = static_cast<double>(2197)/static_cast<double>(4104);
    double b45 = static_cast<double>(-1)/static_cast<double>(5);
    int idx = blockIdx.x * blockDim.x + threadIdx.x;
    y4[idx] = concs[idx] + b41 * k1s[idx] + b42 * k2s[idx] + b43 * k3s[idx] + b44 * k4s[idx] + b45 * k5s[idx];
}
//takeFifthOrderStep is going to overwrite the old array of concentrations with the new array of concentrations.  As of now, this will be the 5th order step.  Another function can be d
//defined that will take a fourth order step if that is interesting for any reason.  This kernel is meant to be launched with as many threads as there are discrete concentrations
//to be tracked.
//Store b values in register? Constants?
__global__ void takeFifthOrderStep(double* y5, double* concs, double* k1s, double* k2s, double* k3s, double* k4s, double* k5s, double* k6s)
{
    double b51 = static_cast<double>(16)/static_cast<double>(135);
    double b52 = static_cast<double>(0);
    double b53 = static_cast<double>(6656)/static_cast<double>(12825);
    double b54 = static_cast<double>(28561)/static_cast<double>(56430);
    double b55 = static_cast<double>(-9)/static_cast<double>(50);
    double b56 = static_cast<double>(2)/static_cast<double>(55);
    int idx = blockIdx.x * blockDim.x + threadIdx.x;
    y5[idx] = concs[idx] + b51 * k1s[idx] + b52 * k2s[idx] + b53 * k3s[idx] + b54 * k4s[idx] + b55 * k5s[idx] + b56 * k6s[idx];
}
//storeConcs takes the current array of concentrations and stores it in the cId'th column of the 2D concStorage array
//pitch = memory size of a row
__global__ void storeConcs(double* cS, size_t pitch, double* concs, int cId)
{
    int tIdx = threadIdx.x;
    //cS is basically the memory address of the first element of the flattened (1D) 2D array.
    double* row = (double*)((char*)cS + tIdx * pitch);
    row[cId] = concs[tIdx];
}
//Perhaps I can optimize by using shared memory to hold conc values.
__global__ void calcK(double* concs, double* ks, double* dt)
{
    int idx = blockIdx.x * blockDim.x + threadIdx.x;
    ks[idx]=(*dt)*flux(idx, concs);
}
//Adds two arrays (a + b) element by element and stores the result in array c.
__global__ void arrAdd(double* a, double* b, double* c)
{                                                 
    int idx = blockIdx.x * blockDim.x + threadIdx.x;
    c[idx]=a[idx]+b[idx];
}
//Subtracts two arrays (a - b) element by element and stores the result in array c.
__global__ void arrSub(double* a, double* b, double* c)
{                                                 
    int idx = blockIdx.x * blockDim.x + threadIdx.x;
    c[idx]=a[idx]-b[idx];
}
//Multiplies two arrays (a * b) element by element and stores the result in array c.
__global__ void arrMult(double* a, double* b, double* c)
{
    int idx = blockIdx.x * blockDim.x + threadIdx.x;
    c[idx]=a[idx]*b[idx];
}
//Will find the min of errors array.
__global__ void arrMin(double* errors)
{
    //extern _shared_ double[7];
}
//Initializes an array a to double value b.
__global__ void arrInit(double* a, double b)
{
    int idx = blockIdx.x * blockDim.x + threadIdx.x;
    a[idx]=b;
}
//Copies array b onto array a.
__global__ void arrCopy(double* a, double* b)
{
    int idx = blockIdx.x * blockDim.x + threadIdx.x;
    a[idx]=b[idx];
}
//Placeholder function for the flux calculation.  It will take the size of the oligomer and current concentrations as inputs.
__device__ double flux(int r, double *concs) 
{
    return -concs[r];
}
//This function multiplies a tableau value by the corresponding k array and adds the result to tempsum.  Used to
//add all the a*k terms.
//e.g. setup(a21, temp1, tempsum, k1s, concs, maxlength) => tempsum = a21 * k1
__device__ void setup(double tableauValue, double *temp1, double *tempsum, double *ks, double *concs, int *maxlength) 
{
    //Sets tempsum to tabVal * k
    arrInit<<< 1, *maxlength >>>(temp1, tableauValue);      //Set [temp1] to tableau value, temp1 = a
    cudaDeviceSynchronize();
    arrMult<<< 1, *maxlength >>>(ks, temp1, temp1);         //Multiply tableau value by appropriate [k], temp1 = a*k
    cudaDeviceSynchronize();
    arrAdd<<< 1, *maxlength >>>(tempsum, temp1, tempsum);   //Move tabVal*k to [tempsum], tempsum = tempsum+temp1
    cudaDeviceSynchronize();
    //temp1 = tableauValue * kArray
    //tempsum = current sum (tableauValue * kArray)
}

Edição final Encontrei o meu erro. Foi um erro de digitação calculando o K6. Coloquei asteriscos em torno do termo "k4s" ofensivo, que deve ser um "k5s". Qualquer pessoa que queira um integrador CUDA RKF45 pode usá-lo, pois funciona agora e está na ordem correta.

Number of Points    h       Global Error    Calculated Order
        10          0.1         3.59E-09    
        20          0.05        1.08E-10    5.06E+00
        40          0.025       3.31E-12    5.03E+00
        80          0.0125      1.03E-13    5.01E+00
        160         0.00625     2.20E-15    5.55E+00
        320         0.003125    1.20E-15    8.74E-01
        640         0.0015625   1.40E-15    -2.22E-01
        1280        0.00078125  2.00E-16    2.81E+00

Estou assumindo que você está configurando a tolerância a erros em 1e-12. Você está correto quando um esquema adaptativo aceita o tamanho atual da etapa, ele assume que o esquema de 5ª ordem foi, para todos os efeitos, a resposta "correta". No entanto, isso é apenas quando ele aceita a etapa atual. Se a diferença entre as 4ª e 5ª etapas ordem são muito grande, ele não aceita , quer como bom o suficiente. Para testar um único algoritmo como esse, é necessário executá-lo com várias etapas de tempo e ver a ordem da convergência para verificá-lo.

Agora, por outro lado, uma etapa 0.001com um esquema de 5ª ordem no seu problema de teste (suponho que fosse 1D) deve ser mais do que suficiente para levar o erro quase à máquina epsilon. Meu palpite é que você tem um erro de digitação em algum lugar que reduz seu integrador à 1ª ordem. Como eu disse no parágrafo 1, execute-o com várias etapas de tempo e verifique sua ordem de convergência. Isso informará se o erro que você vê é razoável ou se existe algum erro em seu integrador.

EDITAR

$h$$\frac{h}{2}$

$\frac{E(h)}{E(\frac{h}{2})} = 2^n$

$n=5$